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Sagot :
Sure, let's tackle these questions step-by-step.
### 1. Finding the 16th term of the sequence: 28, 24, 20, 16, 124, 8
Given the sequence: 28, 24, 20, 16, 124, 8.
To find the 16th term, we look at the pattern of the sequence. Based on the terms given, it seems the sequence is an arithmetic sequence where the common difference (d) is decreasing by 4 each step. Let's assume the sequence is consistent with that pattern.
Using the formula for the nth term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = 28 \)[/tex] (the first term)
- [tex]\( n = 16 \)[/tex] (the term position we are trying to find)
- [tex]\( d = -4 \)[/tex] (the common difference)
[tex]\[ a_{16} = 28 + (16-1) \cdot (-4) \][/tex]
[tex]\[ a_{16} = 28 + 15 \cdot (-4) \][/tex]
[tex]\[ a_{16} = 28 - 60 \][/tex]
[tex]\[ a_{16} = -32 \][/tex]
So, the 16th term is [tex]\(-32\)[/tex].
### 2. Finding the ith term of the sequence: 2-3, -6, -9, -12, -15
Given the sequence: 2-3, -6, -9, -12, -15.
This sequence seems to be an arithmetic sequence where the common difference (d) is decreasing by 3. Let's assume we are looking for the 6th term (for this example).
Again, using the formula for the nth term:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = -3 \)[/tex] (the first term)
- [tex]\( n = 6 \)[/tex] (assuming i refers to the 6th term)
- [tex]\( d = -3 \)[/tex] (common difference)
[tex]\[ a_{6} = -3 + (6-1) \cdot (-3) \][/tex]
[tex]\[ a_{6} = -3 + 5 \cdot (-3) \][/tex]
[tex]\[ a_{6} = -3 - 15 \][/tex]
[tex]\[ a_{6} = -18 \][/tex]
So, the ith term (assuming i = 6) is [tex]\(-18\)[/tex].
### 3. Finding the 25+1 term of the sequence: 14, 21, 28, 35, 43
Given the sequence: 14, 21, 28, 35, 43.
This sequence is an arithmetic sequence with a common difference of 7.
Using the formula for the nth term:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = 14 \)[/tex] (the first term)
- [tex]\( n = 25 + 1 = 26 \)[/tex] (since we need the 25+1 th term)
- [tex]\( d = 7 \)[/tex] (common difference)
[tex]\[ a_{26} = 14 + (26-1) \cdot 7 \][/tex]
[tex]\[ a_{26} = 14 + 25 \cdot 7 \][/tex]
[tex]\[ a_{26} = 14 + 175 \][/tex]
[tex]\[ a_{26} = 189 \][/tex]
So, the 26th term is [tex]\(189\)[/tex].
### Summary:
1. The 16th term of the sequence [tex]\(28, 24, 20, 16, 124, 8\)[/tex] is [tex]\(-32\)[/tex].
2. The ith term (assuming i = 6) of the sequence [tex]\(2-3, -6, -9, -12, -15\)[/tex] is [tex]\(-18\)[/tex].
3. The [tex]\(25+1\)[/tex] (26th) term of the sequence [tex]\(14, 21, 28, 35, 43\)[/tex] is [tex]\(189\)[/tex].
### 1. Finding the 16th term of the sequence: 28, 24, 20, 16, 124, 8
Given the sequence: 28, 24, 20, 16, 124, 8.
To find the 16th term, we look at the pattern of the sequence. Based on the terms given, it seems the sequence is an arithmetic sequence where the common difference (d) is decreasing by 4 each step. Let's assume the sequence is consistent with that pattern.
Using the formula for the nth term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = 28 \)[/tex] (the first term)
- [tex]\( n = 16 \)[/tex] (the term position we are trying to find)
- [tex]\( d = -4 \)[/tex] (the common difference)
[tex]\[ a_{16} = 28 + (16-1) \cdot (-4) \][/tex]
[tex]\[ a_{16} = 28 + 15 \cdot (-4) \][/tex]
[tex]\[ a_{16} = 28 - 60 \][/tex]
[tex]\[ a_{16} = -32 \][/tex]
So, the 16th term is [tex]\(-32\)[/tex].
### 2. Finding the ith term of the sequence: 2-3, -6, -9, -12, -15
Given the sequence: 2-3, -6, -9, -12, -15.
This sequence seems to be an arithmetic sequence where the common difference (d) is decreasing by 3. Let's assume we are looking for the 6th term (for this example).
Again, using the formula for the nth term:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = -3 \)[/tex] (the first term)
- [tex]\( n = 6 \)[/tex] (assuming i refers to the 6th term)
- [tex]\( d = -3 \)[/tex] (common difference)
[tex]\[ a_{6} = -3 + (6-1) \cdot (-3) \][/tex]
[tex]\[ a_{6} = -3 + 5 \cdot (-3) \][/tex]
[tex]\[ a_{6} = -3 - 15 \][/tex]
[tex]\[ a_{6} = -18 \][/tex]
So, the ith term (assuming i = 6) is [tex]\(-18\)[/tex].
### 3. Finding the 25+1 term of the sequence: 14, 21, 28, 35, 43
Given the sequence: 14, 21, 28, 35, 43.
This sequence is an arithmetic sequence with a common difference of 7.
Using the formula for the nth term:
[tex]\[ a_n = a_1 + (n-1) \cdot d \][/tex]
Where:
- [tex]\( a_1 = 14 \)[/tex] (the first term)
- [tex]\( n = 25 + 1 = 26 \)[/tex] (since we need the 25+1 th term)
- [tex]\( d = 7 \)[/tex] (common difference)
[tex]\[ a_{26} = 14 + (26-1) \cdot 7 \][/tex]
[tex]\[ a_{26} = 14 + 25 \cdot 7 \][/tex]
[tex]\[ a_{26} = 14 + 175 \][/tex]
[tex]\[ a_{26} = 189 \][/tex]
So, the 26th term is [tex]\(189\)[/tex].
### Summary:
1. The 16th term of the sequence [tex]\(28, 24, 20, 16, 124, 8\)[/tex] is [tex]\(-32\)[/tex].
2. The ith term (assuming i = 6) of the sequence [tex]\(2-3, -6, -9, -12, -15\)[/tex] is [tex]\(-18\)[/tex].
3. The [tex]\(25+1\)[/tex] (26th) term of the sequence [tex]\(14, 21, 28, 35, 43\)[/tex] is [tex]\(189\)[/tex].
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